Answer:
The answer is the first one at the tap
Step-by-step explanation:
I am not sure what your problem here is.
you understand the inequality signs ?
anyway, to get
6×f(-2) + 3×g(1)
we can calculate every part of the expression separately, and then combine all the results into one final result.
f(-2)
we look at the definition.
into what category is -2 falling ? the one with x<-2, or the one with x>=-2 ?
is -2 < -2 ? no.
is -2 >= -2 ? yes, because -2 = -2. therefore, it is also >= -2.
so, we have to use
1/3 x³
for x = -2 that is
1/3 × (-2)³ = 1/3 × -8 = -8/3
g(1)
again, we look at the definition.
into what category is 1 falling ? the one with x > 2 ? or the one with x <= 1 ?
is 1 > 2 ? no.
is 1 <= 1 ? yes, because 1=1. therefore it is also <= 1.
so we have to use
2×|x - 1| + 3
for x = 1 we get
2×0 + 3 = 3
6×f(-2) = 6 × -8/3 = 2× -8 = -16
3×g(1) = 3× 3 = 9
and so in total we get
6×f(-2) + 3×g(1) = -16 + 9 = -7
Answer:
Step-by-step explanation:
3(a+3)-6=21
<=>3a+9-6=21
<=>3a+3=21
<=>3a=18
<=>a=6
Answer:
7.
Solution given;
male=15
female=27
1st term=5*3
2nd term=3*3*3
now
Highest common factor=3
So
<u>The</u><u> </u><u>maximum</u><u> </u><u>number</u><u> </u><u>of</u><u> </u><u>groups</u><u> </u><u>that</u><u> </u><u>the</u><u> </u><u>teacher</u><u> </u><u>can</u><u> </u><u>make</u><u> </u><u>is</u><u> </u><u>3</u><u>.</u>
<u>and</u><u> </u><u>each</u><u> </u><u>team</u><u> </u><u>contains</u><u> </u><u>5</u><u> </u><u>male and</u><u> </u><u>9</u><u> </u><u>female</u><u>.</u>
Answer:
Yes, they do.
Step-by-step explanation:
They both pass through the same points of -2 on the x and y axis.